Solution:

Let us solve this problem with variables to find the general nature. Let us assume that there are x black chocolates and y white chocolates. Three things can happen when you take out two chocolates.
1) You take out two black chocolates.
Then, you will be replacing one black chocolate and so the jar will have x+1 black chocolates and y-2 white chocolates.
2) You take out two white chocolates.
Then, you will be replacing 1 black chocolate and so the jar will have x-1 black chocolates and y white chocolates.
3) You take out 1 black chocolate and 1 white chocolate. Then, you will be replacing a white chocolate and so the jar will have x-1 black chocolates and y white chocolates.
Thus at every step, you are wither removing a black chocolate or replacing two white chocolates by one black chocolate irreversibly till there is just one chocolate left. Therefore each white chocolate is equivalent to two black chocolates.

This, if the number of white chocolates are odd in number, the last one remaining will be white chocolate and if the number of white chocolates are even in number, the last one remaining will be black chocolate.

Solution:

Since the knife was in her hand which implies that she was planning to kill her husband "Inspector Montalbano".

Solution:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years

Solution:

The Password.

Explanation:
Clues 2 and 5 are correct and rest are lies.

Solution:

It can be solved as below.

Solution:

Since every bacterium divides into two in two minutes and we know the fact that the dish was half filled with bacteria at 3 pm, after exactly 2 minutes, all will divide into two and will fill the petri dish. Thus, the petri dish will become full at 3:02 pm.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

A Question

Solution:

zero

I had 3 green socks and 1 Pink sock.

=> chances of pair of green socks = 1/2
chances of pair of mixed socks = 1/2
chances of pair of green socks = NULL

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.